In one line
The impulse of the net force on a particle over a time interval equals the change in its linear momentum.
Explained simply
Catching a cricket ball, you pull your hands back. The ball's change in momentum is fixed, but spreading it over a longer time reduces the force on your hands. That trade-off between force and time is exactly the impulse-momentum principle.
Technically, integrating \( \Sigma \vec{F} = m\,d\vec{v}/dt \) over time: \( \int_{t_1}^{t_2} \Sigma \vec{F}\,dt = m\vec{v}_2 - m\vec{v}_1 \). The left side is the impulse; mv is linear momentum.
How it works
- Draw the FBD and the momentum 'before' and 'after' diagrams.
- Choose a positive direction.
- Compute impulse: for constant force, \( F\Delta t \); for variable force F(t), \( \int F\,dt \) (area under the F-t graph).
- Write \( mv_1 + \Sigma \int F\,dt = mv_2 \) along each axis.
- Solve for the unknown velocity, time or average force.
- Useful when force and time are involved and acceleration is not needed.
Key formula
\( m v_1 + \int_{t_1}^{t_2} \Sigma F\,dt = m v_2 \). Momentum in kg·m/s; impulse in N·s (equal units). Average force \( F_{avg} = \Delta(mv)/\Delta t \).
Quick example
A 0.15 kg ball arrives at 30 m/s and is hit straight back at 40 m/s; contact lasts 0.01 s. Taking the return direction positive: impulse \( = 0.15(40 - (-30)) = 10.5 \) N·s. Average force \( = 10.5/0.01 = 1050 \) N.
Common mistakes
- Ignoring the sign change when velocity reverses (giving 1.5 N·s instead of 10.5 N·s).
- Leaving out the weight or friction when they act over a long time interval.
- Using impulse-momentum in a direction where an unknown force acts but not accounting for it.
How MU asks it
Theory 3 marks: 'State the principle of impulse and momentum'. Numerical 5 to 6 marks: a block pushed by a time-varying force, a ball hit by a bat, a body sliding on a rough plane for a given time. State the principle, write the equation along the motion, solve with units.